Using the binomial theorem, Find the first three terms in the expansion of (x-(1/x))^5.
My solution:
C(5,0)x5 + C(5,1)(x4)(-1/x) + C(5,2)(x3)(-1/x)2
x5 + (5)(x4)(-1/x) + 10(x3)(-1/x)2
x5 + (-5x4/x) + 10(x3)(1/x2)
x5 + (-5x3) + 10(x)
x5 -5x3 + 10x
The first three terms are x5, -5x3, and 10x, respectably.
Is this right? Thanks a lot.
what is the answer ?
Yes, your solution is correct. You have correctly applied the binomial theorem to expand (x - 1/x)^5.
To find the first three terms in the expansion, you have used the formula for the binomial coefficients (nCr) and the powers of x and -1/x.
The general form of the binomial theorem is:
(x + y)^n = C(n,0)x^n y^0 + C(n,1)x^(n-1) y^1 + C(n,2)x^(n-2) y^2 + ... + C(n,n)x^0 y^n
In this case, x is the variable and -(1/x) is the constant term. So, when you substitute x and -(1/x) into the binomial theorem, it becomes:
(x - 1/x)^5 = C(5,0)x^5 (-1/x)^0 + C(5,1)x^4 (-1/x)^1 + C(5,2)x^3 (-1/x)^2 + ... + C(5,5)x^0 (-1/x)^5
Simplifying each term using the binomial coefficient formula:
C(5,0) = 1, C(5,1) = 5, C(5,2) = 10, C(5,3) = 10, C(5,4) = 5, C(5,5) = 1
Substituting these values, you get:
(x - 1/x)^5 = x^5 + 5x^4 (-1/x) + 10x^3 (-1/x)^2 + ... + (-1/x)^5
Now, simplifying each term:
(-1/x)^2 = 1/x^2
(-1/x)^3 = -1/x^3
(-1/x)^4 = 1/x^4
(-1/x)^5 = -1/x^5
Substituting these simplified terms, you get:
(x - 1/x)^5 = x^5 + (-5x^3) + 10x
Therefore, the first three terms in the expansion are x^5, -5x^3, and 10x, respectively.
Well done on solving this using the binomial theorem!